Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{10z - 10}{z^2 + 4z - 5} \times \dfrac{z + 5}{-7z - 7} $
Answer: First factor the quadratic. $a = \dfrac{10z - 10}{(z + 5)(z - 1)} \times \dfrac{z + 5}{-7z - 7} $ Then factor out any other terms. $a = \dfrac{10(z - 1)}{(z + 5)(z - 1)} \times \dfrac{z + 5}{-7(z + 1)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ 10(z - 1) \times (z + 5) } { (z + 5)(z - 1) \times -7(z + 1) } $ $a = \dfrac{ 10(z - 1)(z + 5)}{ -7(z + 5)(z - 1)(z + 1)} $ Notice that $(z - 1)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ 10(z - 1)\cancel{(z + 5)}}{ -7\cancel{(z + 5)}(z - 1)(z + 1)} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $a = \dfrac{ 10\cancel{(z - 1)}\cancel{(z + 5)}}{ -7\cancel{(z + 5)}\cancel{(z - 1)}(z + 1)} $ We are dividing by $z - 1$ , so $z - 1 \neq 0$ Therefore, $z \neq 1$ $a = \dfrac{10}{-7(z + 1)} $ $a = \dfrac{-10}{7(z + 1)} ; \space z \neq -5 ; \space z \neq 1 $